Spring-mass system with two springs and three masses

Question

I'm trying to solve a system of springs and masses that is confusing me. First, the balls are all lined up linearly. Secondly, the ball in the middle has a smaller mass $m$ while the first and last balls have a larger mass of $M$. The larger balls are each connected to the middle ball with a spring with a spring constant of $k$. They are assumed to move in the right direction and there is no additional external force.

     x_1      x_2       x_3
      *--------*---------*           ------> x
      M    k   m    k    M

I'm trying to solve this system using the eigenvalue concept, but I'm having trouble. I've dealt with a two mass spring system, but never a three mass, so I'm a bit confused.

From what I gather, I just set the $\ddot{x}$ to $F/m$ and set up the matrix, much like for a two mass system. But I don't know how to do that with the third mass. Like for example, for $\ddot{x}_1$, I think that we have

$$0 -(k/m)(x_1-x_2) -(k/M)(x_2-x_3) $$

and for $\ddot{x}_2$

$$ 0 -(k/m)(x_2-x_1) -(k/M)(x_2-x_3)$$

and for $\ddot{x}_3$

$$ 0+0-(k/M)(x_3-x_2)$$

but I really don't know if I'm right. I highly doubt it. So if some kind soul could tell me what I'm doing wrong/right and point me in the next direction, I would be forever grateful.

Answer 1

OK. You must use the Newton's second law $F=ma$ for each mass. But you also need to specify your system of reference. For example, lets consider our origin in the rest position of the mass $M$ in the left. Respect that system of reference we can write the forces acting upon each mass. The first mass undergoes two forces due to the first spring and to the movements of the two masses 1 and 2. That's why we can write $$F_1\equiv M\frac{d^2x_1}{dt^2}= kx_1+k(x_2-l)$$ Where $x_1$ and $x_2$ means the position of the first $M$ and second $m$ mass with respect our origin. Here $l$ represents the rest length of the springs. Notice that if $x_1=0$ and $x_2=l$, then there is no force upon the left mass $M$. I understand that the problem can be difficult, but in order to understand and clarify the ideas you should make a diagram showing the forces acting upon each mass when vary the positions of each of the three masses.

Answer 2

I made a typing mistake yesterday. The correct period of oscillation should be $$ T=2\pi\frac{(2Mm)^{0.5}}{{(m+2M)^{0.5}}{(2k)^{0.5}}}. $$ It can be derived by applying NLMs & constraint relations & conservation of energy to finding time period in oscillatory motion. Now, my answer is completely correct. I have also varified it in a numerical of Irodov. It will be useful for you to go through that sum in Irodov in the chapter of oscillations. (In my edition of Irodov this sum is in page 175. I think it will be in the same page in all editions.)

Answer 3

Here time period of oscillation would be $$2 \pi \frac{ \sqrt{m+2M} \sqrt{2k}}{\sqrt{2mM}} $$ it can be derived by applying NLMs & constraight relations & energy conservation method of finding time period in oscillatory motion.

Answer 4

The object on the left at $x_1$ is only connected to the small mass so the 'equation of motion' should be,

$$ M \, \ddot{x}_1 = k(x_2 - x_1) = F, $$

The sign of the force experienced by the mass at $x_1$ must be positive because the spring is pulling the mass along the $x$-axis to the right.

You are mixing the masses up as well. The other equations of motion should be,

$$ m \,\ddot{x}_2 = -k(x_2 - x_1) + k(x_3 - x_2)\\ M \,\ddot{x}_3 = -k(x_3 - x_2). $$

The next thing you should do is to try and divide the masses over on the right side of the equations and write them as a matrix equation which in general can be written as $$ \ddot{\mathbf{x}} = \mathbf{A}\mathbf{x} $$ where $\mathbf{A}$ is the system-matrix you should derive from the equations above. The eigenvalues of the system-matrix corresponds to the eigenmode frequencies squared as you can see for yourself by assuming oscillatory solutions of the form $\mathbf{x}(t) = \mathbf{x}_0\cos(\omega t)$ since the second derivative above yield two factors of $\omega$. But this is your job!